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can someone please help. A sucrose solution is prepared to a final concentration of 0.170M . Convert this value into terms of g/L, molality, and mass % (molecular weight, MWsucrose = 342.296g/mol ; density, ρsol′n = 1.02g/mL ; mass of water, mwat = 961.8g ). Note that the mass of solute is included in the density of the solution,

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ChemistryHelper2024
From Molarity: we have 0.17 mole sucrose in 1 Liter solution
1) To convert it into g/L we have to multiply moles of sucrose by its molar mass:
= 0.17 mole sucrose x [tex] \frac{342.296 g sucrose}{1 mole sucrose} [/tex] = 58.2 g/L
2) Molality =  number of moles of solute / mass in kg of solvent = [tex] \frac{0.17 mole solute}{0.9618 kg solvent} [/tex] = 0.177 mole / kg
3) mass % = mass of solute / mass of solution x 100 
mass of solution = 1020 g from its density
mass of solute = 1020 - 961.8 = 58.2 g solute
mass % = [tex] \frac{58.2 g solute}{1020 g solution} [/tex] x 100 = 5.7 %

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